Question

Assume that a resting human on a space capsule uses 14.8 L O2/hr. Assume that all of their O2 consumption is used in the following rxn.: C6H12O6 +6O2 --> 6CO2 +6H2O. What mass of LiOH would be required to purify the air (quantitatively remove the CO2 produced by the astronauts) in the space capsule for 4.7 days for 2 humans? Assume that the cabin of the space capsule has pressure of 0.97 atm and a temperature of 4.5 C. Show all calculations. You must write a balanced chemical equation for the neutralization of CO2 by LiOH.

Answer 1

Answer 2

 Calculation:

1. Calculate Total O₂ Consumed:

• Given:

• O₂ consumption per human = 14.8 L/hr

• Number of humans = 2

• Time = 4.7 days = 4.7 × 24 = 112.8 hours

• Total O₂ consumed:

  $$14.8\text{L/hr/human}\times 2\text{humans}\times 112.8\text{hr}=3,338.88\text{L O₂}$$

2. Relate O₂ to CO₂ Produced:

From the balanced reaction:

$$C_6{H}_{12}{O}_6+6O_2\to 6C{O}_2+6H_{2}O$$

• Mole Ratio:
  6 moles O₂ → 6 moles CO₂ ⇒ 1:1 ratio

• Total CO₂ produced = Total O₂ consumed = 3,338.88 L

3. Convert CO₂ Volume to Moles (Ideal Gas Law):

• Given:

• Pressure (P) = 0.97 atm

• Temperature (T) = 4.5°C = 277.65 K

• Volume (V) = 3,338.88 L

• Gas constant (R) = 0.0821 L·atm/(K·mol)

• Moles of CO₂:

  $$n=\frac{PV}{RT}=\frac{0.97\times 3,338.88}{0.0821\times 277.65}\approx 142.3\text{moles CO₂}$$

4. Determine LiOH Required for CO₂ Neutralization:

Balanced reaction for CO₂ and LiOH:

$$2LiOH+C{O}_2\to L{i}_{2}C{O}_3+H_{2}O$$

• Mole Ratio:
  2 moles LiOH : 1 mole CO₂

• Moles of LiOH needed:

  $$142.3\text{moles CO₂}\times 2=284.6\text{moles LiOH}$$

• Molar Mass of LiOH:
  Li (6.94) + O (16) + H (1.01) = 23.95 g/mol

• Mass of LiOH:

  $$284.6\text{moles}\times 23.95\text{g/mol}=6,816.17\text{g}\approx 6.82\text{kg}$$

5. Adjust for Partial Pressure (Optional Check):

If CO₂ is part of the cabin air (not pure), recalculate using partial pressure. However, the problem implies quantitative removal, so the above calculation stands.

Answer:
To purify the air in the space capsule for 2 humans over 4.7 days, approximately 4.1 kg of LiOH is required.